What is the vertex of the parabola y = (x-4)^2?

Apr 22, 2016

$\left(4 , 0\right)$

Explanation:

Standard form;$\text{ } y = a {x}^{2} + b x + c$
Vertex form;$\text{ } y = a {\left(x + \frac{b}{2 a}\right)}^{2} + k$

So your given equation is in vertex form in that we have:

$\text{ } y = 1 {\left(x - 4\right)}^{2} + 0$

Where ${x}_{\text{vertex}} = \left(- 1\right) \times \frac{b}{2 a} \to \left(- 1\right) \times \left(- 4\right) = + 4$

" "y_("vertex")=k ->0

color(blue)("Vertex "->(x,y)->(4,0)